Q

# Can someone help me with this, - Electrostatics - JEE Main-8

$q_1,q_2,q_3 , q_4$  are point charges located at points as shown in the figure and s is a spherical Gaussian surface of radius R. Which of the following is true according to the Gauss’s law

• Option 1)

$\int_{s}(\vec{E_1}+\vec{E_2}+\vec{E_3})d\vec{A}= \frac{q_1+q_2+q_3}{2\varepsilon _0}$

• Option 2)

$\int_{s}(\vec{E_1}+\vec{E_2}+\vec{E_3})d\vec{A}= \frac{q_1+q_2+q_3}{\varepsilon _0}$

• Option 3)

$\int_{s}(\vec{E_1}+\vec{E_2}+\vec{E_3})d\vec{A}= \frac{q_1+q_2+q_3+q_4}{\varepsilon _0}$

• Option 4)

None of the above

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As we have learned

Gauss's Law -

Total flux linked with a closed surface called Gaussian surface.

Formula:

$\dpi{100} \phi = \oint \vec{E}\cdot d\vec{s}=\frac{Q_{enc}}{\epsilon _{0}}$

- wherein

No need to be a real physical surface.

Qenc - charge enclosed by closed surface.

By using $\int \vec{E}d\vec{A}= Q_{enc}/\varepsilon _0$

Option 1)

$\int_{s}(\vec{E_1}+\vec{E_2}+\vec{E_3})d\vec{A}= \frac{q_1+q_2+q_3}{2\varepsilon _0}$

Option 2)

$\int_{s}(\vec{E_1}+\vec{E_2}+\vec{E_3})d\vec{A}= \frac{q_1+q_2+q_3}{\varepsilon _0}$

Option 3)

$\int_{s}(\vec{E_1}+\vec{E_2}+\vec{E_3})d\vec{A}= \frac{q_1+q_2+q_3+q_4}{\varepsilon _0}$

Option 4)

None of the above

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